Tính các giới hạn sau: a) lim x đến + vô cùng 1 - 2x/ căn bậc hai của x^2+ 1; b) lim x đến + vô cùng ( căn bậc hai của x^2 + x + 2 - x).

Lời giải:

a) \(\mathop {\lim }\limits_{x \to + \infty } \frac{{1 - 2x}}{{\sqrt {{x^2} + 1} }}\)\( = \mathop {\lim }\limits_{x \to + \infty } \frac{{1 - 2x}}{{\sqrt {{x^2}\left( {1 + \frac{1}{{{x^2}}}} \right)} }}\)\( = \mathop {\lim }\limits_{x \to + \infty } \frac{{x\left( {\frac{1}{x} - 2} \right)}}{{x\sqrt {1 + \frac{1}{{{x^2}}}} }}\)\( = \mathop {\lim }\limits_{x \to + \infty } \frac{{\frac{1}{x} - 2}}{{\sqrt {1 + \frac{1}{{{x^2}}}} }} = \frac{{ - 2}}{{\sqrt 1 }} =  - 2\).

b) Ta có: \(\sqrt {{x^2} + x + 2} - x = \frac{{{{\left( {\sqrt {{x^2} + x + 2} } \right)}^2} - {x^2}}}{{\sqrt {{x^2} + x + 2} + x}}\)\( = \frac{{x + 2}}{{\sqrt {{x^2} + x + 2} + x}}\)

Do đó, \(\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + x + 2} - x} \right)\) \( = \mathop {\lim }\limits_{x \to + \infty } \frac{{x + 2}}{{\sqrt {{x^2} + x + 2} + x}}\)

\( = \mathop {\lim }\limits_{x \to + \infty } \frac{{x + 2}}{{\sqrt {{x^2}\left( {1 + \frac{1}{x} + \frac{2}{{{x^2}}}} \right)} + x}}\)\( = \mathop {\lim }\limits_{x \to + \infty } \frac{{x + 2}}{{x\sqrt {1 + \frac{1}{x} + \frac{2}{{{x^2}}}} + x}}\)

\( = \mathop {\lim }\limits_{x \to + \infty } \frac{{x\left( {1 + \frac{2}{x}} \right)}}{{x\left( {\sqrt {1 + \frac{1}{x} + \frac{2}{{{x^2}}}} + 1} \right)}}\)\( = \mathop {\lim }\limits_{x \to + \infty } \frac{{1 + \frac{2}{x}}}{{\sqrt {1 + \frac{1}{x} + \frac{2}{{{x^2}}}} + 1}} = \frac{1}{2}\).